Simplify; express your answer in exponential form. Assume $n\neq 0, r\neq 0$. $\dfrac{{(n^{-2}r^{-2})^{-4}}}{{(n^{-2}r)^{3}}}$
Solution: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(n^{-2}r^{-2})^{-4} = (n^{-2})^{-4}(r^{-2})^{-4}}$ On the left, we have ${n^{-2}}$ to the exponent ${-4}$ . Now ${-2 \times -4 = 8}$ , so ${(n^{-2})^{-4} = n^{8}}$ Apply the ideas above to simplify the equation. $\dfrac{{(n^{-2}r^{-2})^{-4}}}{{(n^{-2}r)^{3}}} = \dfrac{{n^{8}r^{8}}}{{n^{-6}r^{3}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{8}r^{8}}}{{n^{-6}r^{3}}} = \dfrac{{n^{8}}}{{n^{-6}}} \cdot \dfrac{{r^{8}}}{{r^{3}}} = n^{{8} - {(-6)}} \cdot r^{{8} - {3}} = n^{14}r^{5}$